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Applications of the Heisenberg's uncertainty principle.

Applications of the Heisenberg's uncertainty principle;

The applications of Heisenberg's uncertainty principle are given below;

1) Non-Existence of electron in the nucleus.

We know that the radius of the nucleus of any atom is of the order of 10⁻¹⁴m, so if an electron is confined within the nucleus, the uncertainty in its position must not be greater than 10⁻¹⁴m.
According to uncertainty principle,
∆x.∆p ≈ ℏ/2
Where ∆x = uncertainty in the position.
∆p = uncertainty in the momentum.
And ℏ = h/2π
Since ∆x=r=10⁻¹⁴
∴ ∆p≈h/4π∆x = 6.62×10⁻³⁴/4×3.14× 10⁻¹⁴
        =5.27×10⁻²¹ kgms⁻¹
If this is the uncertainty in the momentum of the electron, the momentum of the electron must be at least comparable with its magnitude, i.e
p ≈ 5.27×10⁻²¹ kgms⁻¹
The kinetic energy of the electron of mass m(=9.1×10⁻³¹kg) is given by
E = p²/2m ≈ [5.27× 10⁻²¹]²/ 2×9.1×10⁻³¹J
E = [5.27×10⁻²¹]²/2×9.1×10⁻³¹×1.6×10⁻¹³ MeV
   ≈ 95.5 MeV [∴1MeV=1.6×10⁻¹³J]
It means that if the electron exist inside the nucleus, their kinetic energy must be of the order of 95.5 MeV. But experimental observations reveals that no electron in the atom can have energy greater than 4MeV. Therefore, electrons do not exist in the nucleus.

2) Minimum Energy of particle in a box:

Let us consider a particle in a one dimensional box of length l. then the uncertainty in position is ∆x=l
∴ ∆p≈ ℏ/2∆x = ℏ/2l
=> p≈ ℏ/2l
We know that energy E is given by 
E = p²/2m = ℏ²/2×4l²m = ℏ²/8l²m
Which is the minimum energy of particle in a box.

3) Energy and the radius of the Bohr's first orbit:

Let 'r' be the radius of the atomic orbit. Then ∆x=r.
∆p≈ p ≈ ℏ/2∆x = ℏ/2r
We know that kinetic energy T is
T = p²/2m
=> T =  ℏ²/8r²m ≈  ℏ²/2mr²
Also potential energy V is
 V = -e²/4πε₀r
∴ E= T+V=   ℏ²/2mr²-e²/4πε₀r
Which is the required energy of Bohr's first orbit.
Now 
dE/dr=0
=> d/dr[  ℏ²/2mr²-e²/4πε₀r]=0
=>  ℏ²/2m(-2r⁻³)  -e²/4πε₀(-r⁻²)=0
=> -ℏ²/mr³+e²/4πε₀r²=0
=> e²/4πε₀r² = ℏ²/mr³
=> r = 4πε₀ℏ²/me²
Which is the required radius of Bohr's first orbit.

4) Binding Energy of an electron in an atom.

Let an electron is revolving around the nucleus of an atom in an orbit of radius r.The uncertainty in the position of the electron ∆x is equal to the radius of the orbit.
∴ ∆x = r
According to uncertainty principle, ∆p∆x=ℏ/2
∴ ∆p=ℏ/2r
Hence minimum value of the momentum of the electron in its orbits,
p=∆p=ℏ/2r
Taking r to be of the order of 10⁻¹⁰m, the value of momentum is
p=ℏ/2r=h/4πr= 6.6×10⁻³⁴/4π×10⁻¹⁰ = 
5.2× 10⁻²⁵ kgms⁻¹
The kinetic energy of the electron
K=p²/2m = (5.2× 10⁻²⁵ )²/2×9.1×10⁻³¹ ≈ 1.5×10⁻¹⁹J = 1eV
The potential energy of the electron in the field of the nucleus with atomic number Z is
V = -Ze²/4πε₀r = -Z(1.6×10⁻¹⁹)²×9×10⁹/10⁻¹⁰
   =-Z×2.304×10⁻¹⁸J = -Z×14.4eV
Therefore, the total binding energy is
E=K+V = (1-14.4Z) eV
Eg; For hydrogen Z=1
∴ E= 1-14.4 = -13.4eV
For helium, Z=2
∴E= 1-28.8 = -27.8eV

Fill in the blanks

1) The radius of the nucleus of any atom is of the order of ______ m.
2) Experimental observations reveals that no electron in the atom can have energy greater than ____ MeV.
3) Electrons do not exist in the _______ .
4) The minimum value of energy possessed by a particle in the box is called the __________ energy.
Ans) 1) The radius of the nucleus of any atom is of the order of  10⁻¹⁴ m.
2) Experimental observations reveals that no electron in the atom can have energy greater than 4 MeV.
3) Electrons do not exist in the nucleus .
4) The minimum value of energy possessed by a particle in the box is called the zero point energy.

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